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CC'' and AA'' cut BB'' at X and Y respectively, GE cut HF at Z.considering BAB'' cut by C''C and C'C, and BB''C cut by AA' and AA'', and by Menelau's Theorem,we get BX:XH:HG:GY:YB''=60:24:21:15:20 and HD//GE, so are the others,so ABC, GME and DFH are similar triangles and so are the other 6.considering HYF cut by GE and by Menelau's Theorem, we get FZ:ZH=1:3 and so are the other 2.as we have the length ratio of 4 similar triangles, we get also their area ratio.Let the smallest similar triangle have area S_2, thenS_1=3S_2+[GME]=3(1/4)^2[DFH]+[GME]=3(1/4)^2(84/(140*3))^2S+(105/(140*3))^2S=7/100S

Solution to problem 125.Let K be the center of gravity of ABC. It was proved in problem 122 that M and F are on the median BN, as well as E and H are on the median CP. As proved in problem 123, GE is parallel to AC, GM is parallel to AB, and ME is parallel to BC. Then the six triangles that form the star points are similar to triangle ABC. Let GE and KF meet at point U.We know that KN/BN = 1/3 (1). In problem 121 we have seen that KF/KN = 2/5 (2), and KE/KC = 1/4. As KUE and KNC are similar triangles, then KU/KN = 1/4 (3).From (1) and (2), KF/BN = 2/15 (4). From (2) and (3), KF/KU = 8/5, or UF/KF = 3/8 (5).From (4) and (5), UF/BN = 1/20. This is the factor of similarity of the star point triangle of vertex F and ABC. Similarly the star point triangles with vertices D and H have this same factor of similarity with ABC.Taking S(D), S(F) and S(H) for the areas of the corresponding star point triangles, thenS(D) = S(F) = S(H) = S/400 and S(D) + S(F) + S(H) = 3S/400. From problem 123, S(EGM) = S/16.Hence S1 = S(EGM) + S(D) + S(F) + S(H) = S/16 + 3S/400 = 7S/100.